A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the spaceship = 1000 kg, mass of the sun `= 2 xx 10^(30) kg,` mass of mars `= 6.4 xx 10^(23) kg,` radius of mars = 3395 km, radius of the orbit of mars `= 2.28 xx 10^(8) km, G = 6.67 xx 10^(-11) Nm^(2) kg^( 2)`

Explanation

`implies` Mass of sun M `= 2 xx 10^(30) kg `
Mass of planet mars m `= 6.4 xx 10^(23) kg`
Mass of spaceship `Deltam = 1000 kg `
Orbital radius of planet mars `r_0 = 2.28 xx 10^(11) m `
Radius of mars `r = 3.395 xx 10^(6 )m`
Suppose, orbital velocity of planet mars is v, M
` :. ` Centripetal force = Gravitational force of sun
`(mv^2)/r_0 = (GMm)/(r_0^2) " "....(1)`
Since the velocoity of spaceship is same as the velocity of mars, hence the kinetic energy of spaceship [ `:.` Spaceship is stationary at mars]
`K =1/2 Deltamv^2`
`=1/2 Deltam(GM)/r_0 " " [ :. ` From equ (1) ]
`= (GMDeltam)/(2r_0)`
`= (6.67 xx10^(-17)( xx 2xx 10^(30) xx 10^3))/(2 xx 2.28 xx 10^(11))`
`= (13.34 xx10^(22))/(4.56 xx 10^(11))`
`:. K = 2.9254xx10^(11)J`
{Total potential energy of spaceship V} = {Potential energy due to gravitation field of mars} + {Potential energy due to gravitation field of sun}
`:. V = -(GMDeltam)/r - (GMDeltam)/r_0`
`=-GDeltam[m/r+M/r_0]`
`=-6.67 xx10^(-11)xx10^(3)[(6.4 xx10^(23))/(3.395xx10^6)+(2xx10^(30))/(2.28xx10^11)]`
`=- 6.67xx10^(8) (1.885 xx10^(17) + 87.719 xx10^(17))`
`=- 6.67 xx 10^(-8) xx 89.604 xx10^(17)`
`=- 59.7 . 658 xx10^(9) `
`:. V = - 5.977 xx 10^(11) J`
`:.` Total energy of spaceship,
E = K + V
`= 2.925 xx 10^(11) - 5.977 x 10^(11)`
` = -3.072 xx 10^(11)`
`:. E ~~ - 3.1 xx 10^(11) J`
Negative sign indicates that spaceship is in the bound state of solar system. Hence, to escape from solar system necessary energy required is  ` 3.1 xx 10^(11 ) J`.